Selenium get http request ajax url after button click - ajax

Question

I need to get the http request URL when a button is clicked in a webpage programatically.
I am using selenium to trace the button and I am performing click on the button. on click of the button it makes a http request and the same can be traced in the network tab of the browser.
How can i get the request URL programatically once I trigger the button click using selenium.
Any other tools or libraries that I can use to achieve the same functions is also ok for me. I just need to be able to get the URL after button click programatically. This is a dynamic URL which changes periodically and the objective is to automate the download process through code.
Thanks in advance for any help!

Answer 1

You can use JavaScript executor to get network data. Please refer https://stackoverflow.com/a/45859018/5966329
get request/response data from there.
DesiredCapabilities d = DesiredCapabilities.chrome();
LoggingPreferences logPrefs = new LoggingPreferences();
logPrefs.enable(LogType.PERFORMANCE, Level.ALL);
d.setCapability(CapabilityType.LOGGING_PREFS, logPrefs);
WebDriver driver = new ChromeDriver(d);
driver.get("https://www.google.co.in/");
driver.manage().timeouts().implicitlyWait(15, TimeUnit.SECONDS);
LogEntries les = driver.manage().logs().get(LogType.PERFORMANCE);
for (LogEntry le : les) {
System.out.println(le.getMessage());
}
Python Equivalent:
from selenium import webdriver
from selenium.webdriver.common.desired_capabilities import
DesiredCapabilities
caps = DesiredCapabilities.CHROME
caps['goog:loggingPrefs'] = {'performance': 'ALL'}
driver = webdriver.Chrome(desired_capabilities=caps)
driver.get('https://stackoverflow.com')
for entry in driver.get_log('performance'):
print(entry)
Please refer thread for more info python